dgrozev wrote: Find all functions $f: \mathbb{Q}^+ \to \mathbb{R}^+ $ with the property: \[f(xy)=f(x+y)(f(x)+f(y)) \,,\, \forall x,y \in \mathbb{Q}^+\] Let $P(x,y)$ be the assertion $f(xy)=f(x+y)(f(x)+f(y))$ Let $a=f(1)$ $P(1,1)$ $\implies$ $f(2)=\frac 12$ $P(2,1)$ $\implies$ $f(3)=\frac 1{2a+1}$ $P(3,1)$ $\implies$ $f(4)=\frac 1{2a^2+a+1}$ $P(4,1)$ $\implies$ $f(5)=\frac 1{2a^3+a^2+a+1}$ $P(2,3)$ $\implies$ $f(6)=f(5)(f(2)+f(3))$ $P(5,1)$ $\implies$ $f(5)=f(6)(f(1)+f(5)$ and so $(f(2)+f(3))(f(1)+f(5))=1$ Using values got previously in this equation and simplifying, we get $(a-1)(a+1)(2a-1)(2a^2+a+1)=0$ and so $a\in\{\frac 12,1\}$ 1) If $a=1$ ========= $P(n,1)$ $\implies$ $\frac 1{f(n+1)}=\frac 1{f(n)}+1$ and a simple induction gives $f(n)=\frac 1n$ $P(x+n,1)$ $\implies$ $\frac 1{f(x+n+1}=1+\frac 1{f(x+n)}$ and a simple induction gives then $f(x+n)=\frac{f(x)}{nf(x)+1}$ $P(x,n)$ $\implies$ $f(nx)=\frac{f(x)}n$ and so $f(\frac pq)=\frac qp$ and so : $\boxed{\text{S1 : }f(x)=\frac 1x\text{ }\forall x\in\mathbb Q^+}$ which indeed is a solution. 2) If $a=\frac 12$ ============== $P(x,1)$ $\implies$ $f(x+1)=\frac{2f(x)}{2f(x)+1}$ and a simple induction implies $f(n)=\frac 12$ This also implies $f(x+2)=\frac {4f(x)}{6f(x)+1}$ and $f(x+4)=\frac{16f(x)}{30f(x)+1}$ Then $P(x,2)$ $\implies$ $f(2x)=\frac{2f(x)(2f(x)+1)}{6f(x)+1}$ and $P(2x,2)$ $\implies$ $f(4x)=\frac{4f(x)(2f(x)+1)(8f(x)^2+10f(x)+1)}{(6f(x)+1)(24f(x)^2+18f(x)+1)}$ But $P(x,4)$ $\implies$ $f(4x)=\frac{8f(x)(2f(x)+1)}{30f(x)+1}$ Equating these two expressions of $f(4x)$, we get $(2f(x)-1)^2(12f(x)+1)=0$ and so : $\boxed{\text{S2 : }f(x)=\frac 12\text{ }\forall x\in\mathbb Q^+}$ which indeed is a solution.

Now let's make this more challenging (and this was the original problem of Nikolai Nikolov): same question with rational numbers replaced with positive real numbers.

Yes, the author of the problem is Nikolay Nikolov. It was given at the competition in the form as it have been posted here. To the extent that I have seen, the more complicated version was: Find all functions $ f: \mathbb{R}^+ \to \mathbb{R}$ with the same property.

Let's find the functions $f:(0,\infty)\to \mathbf{R}$ such that \[f(xy)=f(x+y)(f(x)+f(y)).\] This is the original version of the problem, apparently submitted to IMO last year and rejected since considered too hard. My solution is pretty long. Step 1: Let $Z$ be the set of zeros of $f$ and suppose that $S$ is nonempty and $f$ is not the zero map. Observe that: 1) If $x,y\in S$, then $xy\in S$. 2) If $x\in S$, then $u\in S$ for all $u\in (0, x^2/4]$. Indeed, for any such $u$ we can find $0<y<x$ such that $ u=y(x-y)$ and then $f(u)=f(x)(f(y)+f(x-y))=0$. Hence, if $x\in S$ and $n\geq 1$, then $(0, x^{2n}/4]\subset S$. If $x>1$, then by choosing $n$ very large we obtain that $(0, A]\subset S$ for all $A>0$, so $f=0$, contradiction. Hence $S\subset (0, 1]$. Observation 2) above shows that we can find $y_1<y_2$ in $S$. Set $x=y_2/y_1>1$, so that $f(y_2)=f(xy_1)=f(x+y_1)f(x)\ne 0$ (because $x>1$ and $x+y_1>1$), a contradiction. This shows that if $f$ vanishes at some point, then $f$ is the zero map. From now on I will suppose that $f$ is not the zero map. Since $f(1)=2f(2) f(1)$, we have $f(2)=1/2$. Step 2. I claim that $f(1)\in \{1, -1, 1/2\}$. Let $\alpha=f(1)\ne 0$ and suppose that $\alpha\ne 1, 1/2$. Set $x_n=1/f(n)$. Taking $x=n, y=1$, we obtain $f(n)=f(n+1)(f(n)+\alpha)$, hence $x_{n+1}=1+\alpha x_n$. This gives us explicit values for $x_n$ in terms of $\alpha$. In particular, we find $x_3=1+2\alpha$, $x_5=1+\alpha+\alpha^2+2\alpha^3$ and $x_6=1+\alpha+\alpha^2+\alpha^3+2\alpha^4$. Now, we have \[f(6)=f(2\cdot 3)=f(5)( 1/2+f(3)).\] Writing this in terms of $x_3, x_5, x_6$ and using the above formulae we arrive at the nasty equation \[ 3\alpha^3+\alpha^2+\alpha= 4\alpha^5+1.\] This factors as \[ (\alpha^2-1)(2\alpha-1)(2\alpha^2+\alpha+1)=0\] and finishes the proof of step $2$. Step 3. We consider the case $f(1)=1$. Then $f(x+1)=\frac{f(x)}{f(x)+1}$ (take $y=1$), so $f(x+2)=\frac{f(x)}{2f(x)+1}$. Taking $y=2$ we obtain \[f(2x)=f(x+2)(f(x)+1/2)=\frac{f(x)}{2}.\] Taking $y=x$ and using the previous relation we also obtain $f(x^2)=f(x)^2$. In particular $f(x)>0$ for all $x$. Next, by induction we have $f(x+n)=\frac{f(x)}{nf(x)+1}<\frac{1}{n}$ and $f(n)=1/n$. which shows that $f(x)\leq 1/n$ for $x\geq n$. Hence for all $x\geq 1$ we have $f(x)\leq 1/[x]$. But then for all $x>1$ we have \[f(x)^{2^n}=f(x^{2^n})\leq 1/[x^{2^n}].\] Taking the $2^n$th root and passing to the limit we finally obtain $f(x)\leq 1/x$ for $x\geq 1$. But since $f(2^n x)= f(x)/2^n$, we get $f(x)\leq 1/x$ for all $x\geq 2^{-n}$ and this for all $n$, hence $f(x)\leq 1/x$ for all $x$. Combined with the relation \[ 1=f(1)= f(x+1/x)(f(x)+f(1/x))\] this immediately implies $f(x)=1/x$ for all $x$. So we get another solution, the map $x\to 1/x$, unique solution for which $f(1)=1$. Step 4. We consider the case $f(1)=1/2$. This time $f(x+1)= \frac{f(x)}{f(x)+1/2}$ and by induction we get \[f(x+n)=\frac{2^n f(x)}{2 (2^n-1) f(x)+1},\] in particular $f(n)=1/2$ for positive integers $n$ and $\lim_{n\to\infty} f(x+n)=1/2$. Next, we have $f(nx)= f(x+n)(f(x)+1/2)$, and this tends to $ 1/2 (f(x)+1/2)$ when $n\to\infty$. Hence for all positive integers $m$ we have \[ 1/2 ( f(mx)+1/2)=\lim_{n\to\infty} f(mnx)=\lim_{n\to\infty} f(nx)= 1/2(f(x)+1/2),\] hence $f(mx)=f(x)$ for all $x$ and all positive integers $m$. Making again $m\to\infty$ we obtain $f(x)=\lim_{m} f(mx)= 1/2 (f(x)+1/2)$, hence $f(x)=1/2$ for all $x$. This gives another solution to the problem. Step 5. I consider the case $f(1)=-1$ and I claim that there is no solution in this case. Now, we have $f(x+1)= \frac{f(x)}{f(x)-1}$ and replacing $x$ by $x+1$ we obtain $f(x)=f(x+2)$. This gives in particular $f(4)=f(2)=1/2$ and also \[ f(2x)= f(x+2)(f(x)+1/2)=f(x) (f(x)+1/2), f(4x)=f(x+4)(f(x)+1/2)=f(x) (f(x)+1/2).\] Hence $f(2x)=f(4x)$ for all $x$ and so $f(x)=f(2x)=f(x)(f(x)+1/2)$ for all $x$, which gives $f(x)+1/2=1$ for all $x$, a contradiction.

Yes Harazi you are right , it proposed to the last year IMO , and I don't know why it not selected , As I see this try to define function g(x)= 1/f(x) ....My approach is same as you except for finding f(1) , which I used Geometric series ..... . After this problem this problem was proposed to Brazilian math Olympiads http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=559598&p=3256076&hilit=Brazil+2013#p3256076 and Also this problem proposed to ELMO-2012 (Apparently before the propositions to the IMO). http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728445&sid=e8a07fecc683143cc542e3537478bec5#p2728445

Let $g(x) = \frac{1}{f(x)}$ - then we work with $g(xy)((g(x)+g(y)) = g(x+y)g(x)g(y)$. In particular, $g(x+1) = 1 + \frac{g(x)}{g(1)}$, so with $a=g(1)$ we obtain $g(2) = 2$, $g(3) = 1 + \frac{2}{a}$, $g(4) = 1 + \frac{1}{a} + \frac{2}{a^2}$, $g(5) = 1 + \frac{1}{a} + \frac{1}{a^2} + \frac{2}{a^3}$ and $g(6) = 1 + \frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3} + \frac{2}{a^4}$. On the other hand, $x=3$ with $y=2$ yield $g(6) = \frac{g(2)g(3)g(5)}{g(2) + g(3)} = \frac{2\left(1+\frac{2}{a}\right)\left(1 + \frac{1}{a} + \frac{1}{a^2} + \frac{2}{a^3}\right)}{3+\frac{2}{a}}$. Equating the expressions for $g(6)$ leads to $(a-1)(a-2)(a+1)(a^2+a+2) = 0$, thus (as $a>0$) $a=1$ or $a=2$. If $a=1$, then $g(x+1) = g(x) + 1$, so $g(n) = n$ for all positive integers $n$ by induction. Now setting $y = \frac{x}{x-1}$ (so that $xy=x+y$) yields $g\left(1 + \frac{1}{x-1}\right)g(x) = g\left(1 + \frac{1}{x-1}\right) + g(x)$, thus $g\left(1 + \frac{1}{m}\right) = 1 + \frac{1}{m}$ for any positive integer $m$, which with $g(x+1) = g(x) + 1$ implies $g\left(k + \frac{1}{m}\right) = k + \frac{1}{m}$ for all non-negative integers $k$. Finally, for any positive rational number $\frac{p}{q}$, setting $x=p$ and $y=\frac{1}{q}$ now yields $g\left(\frac{p}{q}\right) = \frac{p}{q}$. Therefore $g(x) = x$ and hence $f(x) = \frac{1}{x}$ for all $x$. If $a=2$, then $g(x+1) = 1 + \frac{g(x)}{2}$, so $g(n) = 2$ for all positive integers $n$ by induction. Now setting $y = \frac{x}{x-1}$ (so that $xy=x+y$) yields $g\left(1 + \frac{1}{x-1}\right)g(x) = g\left(1 + \frac{1}{x-1}\right) + g(x)$, thus $g\left(1 + \frac{1}{m}\right) = 2$ for any positive integer $m$, which with $g(x+1) = 1 + \frac{g(x)}{2}$ implies $g\left(k + \frac{1}{m}\right) = 2$ for all non-negative integers $k$. Finally, for any positive rational number $\frac{p}{q}$, setting $x=p$ and $y=\frac{1}{q}$ now yields $g\left(\frac{p}{q}\right) = 2$. Therefore $g(x) = 2$ and hence $f(x) = \frac{1}{2}$ for all $x$.